Question

1. find the measure of the central angle ∠1 in each of the following figures.
2. the point o is the center of the circle circumscribed about triangle abc, with m∠boc = 120° and m∠aob = 140°, as shown. what is the degree measure of ∠abc? (2007 amc 12b problem, question #3)

a. 35
b. 40
c. 45
d. 50
e. 60

Explanation:

Question 1

Step1: Recall circle - angle property

The sum of central angles in a circle is 360°.

For the first figure

Let the central angle be $\angle1$. We know that the sum of the given angle and $\angle1$ is 360°. Given the other angle is 85°. So, $\angle1=360^{\circ}- 85^{\circ}=275^{\circ}$.

For the second figure

The sum of the given angle and $\angle1$ is 360°. Given the other angle is 280°. So, $\angle1 = 360^{\circ}-280^{\circ}=80^{\circ}$.

For the third figure

The sum of the two given non - $\angle1$ angles and $\angle1$ is 360°. The non - $\angle1$ angles are 120° and 150°. So, $\angle1=360^{\circ}-(120^{\circ}+150^{\circ})=360^{\circ}-270^{\circ}=90^{\circ}$.

Question 2

Step1: Use the central - inscribed angle relationship

The measure of an inscribed angle is half of the measure of the central angle that subtends the same arc.
The central angle corresponding to arc $AC$ is $\angle AOC$. First, find $\angle AOC$: Since the sum of central angles around a point is 360°, $\angle AOC=360^{\circ}-\angle AOB - \angle BOC=360^{\circ}-140^{\circ}-120^{\circ}=100^{\circ}$.
The inscribed angle $\angle ABC$ subtends the same arc $AC$ as the central angle $\angle AOC$. By the inscribed - central angle theorem, $m\angle ABC=\frac{1}{2}m\angle AOC$.
So, $m\angle ABC = 50^{\circ}$.

Answer:

1.

• First figure: 275°
• Second figure: 80°
• Third figure: 90°

1. D. 50